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LeetCode 3600 | Maximize Spanning Tree Stability with Upgrades | Java | Hard | CodeByTushu

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LeetCode 3600 | Maximize Spanning Tree Stability with Upgrades | Java | Hard | CodeByTushu

Description: πŸš€ Welcome Back to CodeByTushu!

In today’s video, we solve LeetCode 3600 β€” Maximize Spanning Tree Stability with Upgrades using Binary Search on Answer and Disjoint Set Union in Java.

We are given an undirected graph containing mandatory and optional edges. Mandatory edges cannot be upgraded, while optional edges can have their strength doubled using at most k upgrades.

The stability of a spanning tree is defined by the minimum strength among all its edges. Our goal is to maximize this stability or return minus one if a valid spanning tree cannot be formed.

We use Binary Search on Answer because the feasibility of a target stability follows a monotonic property. For each candidate value X, we check whether all mandatory edges meet the required strength without forming cycles.

Next, we greedily connect the remaining components using free edges that already meet the required strength. If necessary, we use eligible edges whose doubled strength reaches the target, while staying within the upgrade limit.

We use Disjoint Set Union with path compression to efficiently track connected components. If all vertices become connected, the candidate stability is feasible.

πŸ”₯ Problem Highlights

βœ… Binary Search on Answer using monotonic feasibility
βœ… Checked mandatory edge validity and cycle formation
βœ… Used Disjoint Set Union with path compression and rank optimization
βœ… Prioritized free edges before upgraded edges
βœ… Applied an efficient upper bound based on mandatory edge strengths
βœ… Time Complexity: O((N + M) * alpha(N) * log(max S))
βœ… Auxiliary Space: O(N)

⏱️ TIMESTAMPS

00:00 Intro & Daily Challenge Overview
01:34 Problem Understanding & Mandatory vs Upgrades
03:44 Key Algorithmic Insights: Binary Search & DSU Feasibility
06:05 3-Tier DSU Greedy Algorithm & Range Updates
07:47 Clean Java Solution Walkthrough Line by Line
08:59 Test Cases Walkthrough & Step-by-Step Simulation
10:57 Time Complexity Analysis (O(E log(max S) * alpha(N)) - 30 ms)
11:57 Space Complexity Analysis (O(N) Auxiliary Space - 800 KB)
12:34 Outro & Resources

πŸ“‚ Problem Folder & Detailed Documentation:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/tree/main/solutions/March/12-March/3600-Maximize-Spanning-Tree-Stability-with-Upgrades

β˜• Java Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/March/12-March/3600-Maximize-Spanning-Tree-Stability-with-Upgrades/Solution.java

🐍 Python Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/March/12-March/3600-Maximize-Spanning-Tree-Stability-with-Upgrades/solution.py

⚑ C++ Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/March/12-March/3600-Maximize-Spanning-Tree-Stability-with-Upgrades/solution.cpp

🌐 JavaScript Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/March/12-March/3600-Maximize-Spanning-Tree-Stability-with-Upgrades/solution.js

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πŸ’¬ Question of the Day

Why is it optimal to add all eligible free edges before using upgraded edges during the binary search feasibility check?

1️⃣ Spanning forest matroid properties ensure that prioritizing zero-cost edges minimizes the total number of upgrades needed
2️⃣ It reduces DSU find operations to constant time
3️⃣ Edges meeting the target strength can never form cycles with mandatory edges
4️⃣ It prevents edges from being upgraded more than twice

#LeetCode3600 #LeetCode #MaximizeSpanningTreeStabilityWithUpgrades #Java #JavaDSA #Graph #MinimumSpanningTree #BinarySearch #UnionFind #DSA #CodingInterview #ProblemSolving #LeetCodeHard #DailyLeetCode #JavaProgramming #CodeByTushu

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